Question

Factories $\dfrac{1}{{(x - 3)}} + \dfrac{2}{{(x - 2)}} = \dfrac{8}{x}$

Answer 1

Hint: First, we need to know the concept of LCM,
LCM is the least common multiple, first to find the common multiples among two or more than two numbers and then finding the least among that is the concept.
Using the LCM, we simplify the equation further and try to solve the given problem.
Here x is the unknown variable, so we are going to find the unknown value x, as given in the problem we simplified.
Formula used: The given problem is the quadratic function (at most two degrees) so we make use of the quadratic formula that $\dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$

Complete step by step answer:
From the given that we have the equation in the fraction $\dfrac{1}{{(x - 3)}} + \dfrac{2}{{(x - 2)}} = \dfrac{8}{x}$ (Since the given fractions will be the representation of the number, which are the part of whole numbers and x is the unknown value)
To find the value x, first, we use the least common multiple methods to generalize the given equation as $\dfrac{1}{{(x - 3)}} + \dfrac{2}{{(x - 2)}} = \dfrac{8}{x} \Rightarrow \dfrac{{(x - 2) + 2(x - 3)}}{{(x - 3)(x - 2)}} = \dfrac{8}{x}$this can be obtained from the LCM table as below;
$
  (x - 3)\left| \!{\underline {\,
  {x - 3,x - 2} \,}} \right. \\
  (x - 2)\left| \!{\underline {\,
  {1,x - 2} \,}} \right. \\
 $ On the right side, we don’t need to apply for the LCM because there is only one term.
Now solving further, the LCM converted equation we get, $\dfrac{{(x - 2) + 2(x - 3)}}{{(x - 3)(x - 2)}} = \dfrac{8}{x} \Rightarrow \dfrac{{x - 2 + 2x - 6}}{{(x - 3)(x - 2)}} = \dfrac{8}{x}$
Now cross multiplying the equation we get, $\dfrac{{3x - 8}}{{(x - 3)(x - 2)}} = \dfrac{8}{x} \Rightarrow x(3x - 8) = 8(x - 3)(x - 2)$
Just solving the equation step by step further, $x(3x - 8) = 8(x - 3)(x - 2) \Rightarrow 3{x^2} - 8x = 8{x^2} - 40x + 48$
$3{x^2} - 8x = 8{x^2} - 40x + 48 \Rightarrow 5{x^2} - 32x + 48 = 0$
Now applying the quadratic rule, that $\dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$and here the $a = 5$,$b = - 32$and$c = 48$
Thus, we get, \[\dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}} = \dfrac{{ + 32 \pm \sqrt {{{( - 32)}^2} - 4(5)(48)} }}{{2(5)}}\]
$ \Rightarrow \dfrac{{ + 32 \pm \sqrt {{{( - 32)}^2} - 4(5)(48)} }}{{2(5)}} = \dfrac{{32 \pm \sqrt {1024 - 960} }}{{10}}$
$ \Rightarrow \dfrac{{32 \pm \sqrt {1024 - 960} }}{{10}} = \dfrac{{32 \pm \sqrt {64} }}{{10}} \Rightarrow \dfrac{{32 \pm 8}}{{10}}$
Applying the plus or minus term, we get $\dfrac{{40}}{{10}},\dfrac{{24}}{{10}}$
Thus, we get, $\dfrac{{40}}{{10}},\dfrac{{24}}{{10}} \Rightarrow 4,\dfrac{{12}}{5}$which are the values of the x in the given problem.

Note: For factoring quadratic equation only, we can apply the formula that $\dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$because it has at most two values in the outcome, similarly we don’t need to any formula to find the linear degree operation, because the value itself has the one outcome.
Since the concept of the LCM is to find the smallest number, we used the concept of the LCM to factorize.
There were some other concepts in GCD, the greatest common division, which is to find the common divisor and then choosing which number has the greater value.
If the $\gcd (a,b) = 1$is known as the relatively prime, where a and b are the integers.